Answer:
The 1st and 4th options are correct
I.the oxidized form has a higher affinity for electrons
IV. the greater the tendency for the oxidized form to accept electrons
Explanation:
Half reaction can be described as the oxidation or reduction reaction in a redox reaction.it is In the redox rection there is a change in the oxidation states of Chemical species involved. the oxidized form in the redox has a higher affinity for electrons and the greater the tendency for the oxidized form to accept electrons.
Standard reduction potential which is also referred to as standard cell potential can be described as the potential difference that exist between cathode and anode of the cell. In the standard reduction potential most times the species will be reduced which is usually analysed in a reduction half reaction.
(Standard Hydrogen Electrode) is utilized when determining the Standard reduction or potentials of a chemical specie. this is because of Hydrogen having zero reduction and oxidation potentials, as a result of this a measured potential of any species is compared with that of Hydrogen, the difference helps to know the potential reduction of that particular specie.
If he was 30.8% too low, it means that he was at 69.2% of the boiling point needed. So 50o C is 69.2% of total.
In order to know what 100% is, you can divide the number by it's percentage and then multiply it by a hundred.
So: 50/30.8=1.623
1.623*100=162.3
So the correct boiling point of the liquid he was working with in the lab is 162.3 oC
Answer:
1.263 moles of HF
Explanation:
The balance chemical equation for given single replacement reaction is;
Sn + 2 HF → SnF₂ + H₂
Step 1: <u>Calculate Moles of Tin as;</u>
As we know,
Moles = Mass / A.Mass ----- (1)
Where;
Mass of Tin = 75.0 g
A.Mass of Tin = 118.71 g/mol
Putting values in eq. 1;
Moles = 75.0 g / 118.71 g/mol
Moles = 0.6318 moles of Sn
Step 2: <u>Find out moles of Hydrogen Fluoride as;</u>
According to balance chemical equation,
1 mole of Sn reacted with = 2 moles of HF
So,
0.6318 moles of Sn will react with = X moles of HF
Solving for X,
X = 0.6318 mol × 2 mol / 1 mol
X = 1.263 moles of HF
